Please help! Brainlyest answer!

3 4 5 6

stellarik [79]

C, if you show a picture maybe I can answer it correctly

3 0

3 years ago

Read 2 more answers

(3/5)^2 + 4 x 3 - 2 Please help :)

Molodets [167]

Answer:

$10 \frac{9}{25}$

Step-by-step explanation:

$\bigg( \frac{3}{5} \bigg)^{2} + 4 \times 3 - 2 \\ \\ = \frac{9}{25} + 12 - 2 \\ \\ = \frac{9}{25} + 10 \\ \\ = \frac{9 + 10 \times 25}{25} \\ \\ = \frac{9 + 250}{25} \\ \\ = \frac{259}{25} \\ \\ = 10 \frac{9}{25}$

5 0

2 years ago

suppose a population of 200 crickets double in size every 4 months. How many crickets will there be after 2 years?

tangare [24]

Well 2 years is 24 months. 24 divided by 4 is 6. 200 multiplied by 6 would equal 1200.

The total population of crickets after two years would be 1200.

4 0

3 years ago

I am having trouble with this relative minimum of this equation.

Norma-Jean [14]

Answer:

So the approximate relative minimum is (0.4,-58.5).

Step-by-step explanation:

Ok this is a calculus approach. You have to let me know if you want this done another way.

Here are some rules I'm going to use:

$(f+g)'=f'+g'$ (Sum rule)

$(cf)'=c(f)'$ (Constant multiple rule)

$(x^n)'=nx^{n-1}$ (Power rule)

$(c)'=0$ (Constant rule)

$(x)'=1$ (Slope of y=x is 1)

$y=4x^3+13x^2-12x-56$

$y'=(4x^3+13x^2-12x-56)'$

$y'=(4x^3)'+(13x^2)'-(12x)'-(56)'$

$y'=4(x^3)'+13(x^2)'-12(x)'-0$

$y'=4(3x^2)+13(2x^1)-12(1)$

$y'=12x^2+26x-12$

Now we set y' equal to 0 and solve for the critical numbers.

$12x^2+26x-12=0$

Divide both sides by 2:

$6x^2+13x-6=0$

Compaer $6x^2+13x-6=0$ to $ax^2+bx+c=0$ to determine the values for $a=6,b=13,c=-6$ .

$a=6$

$b=13$

$c=-6$

We are going to use the quadratic formula to solve for our critical numbers, x.

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{-13 \pm \sqrt{13^2-4(6)(-6)}}{2(6)}$

$x=\frac{-13 \pm \sqrt{169+144}}{12}$

$x=\frac{-13 \pm \sqrt{313}}{12}$

Let's separate the choices:

$x=\frac{-13+\sqrt{313}}{12} \text{ or } \frac{-13-\sqrt{313}}{12}$

Let's approximate both of these:

$x=0.3909838 \text{ or } -2.5576505$ .

This is a cubic function with leading coefficient 4 and 4 is positive so we know the left and right behavior of the function. The left hand side goes to negative infinity while the right hand side goes to positive infinity. So the maximum is going to occur at the earlier x while the minimum will occur at the later x.

The relative maximum is at approximately -2.5576505.

So the relative minimum is at approximate 0.3909838.

We could also verify this with more calculus of course.

Let's find the second derivative.

$f(x)=4x^3+13x^2-12x-56$